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Implicit Differentiation

sin x=2t1+t2,...

Question

$sin x = \frac{2 t}{1 + t^{2}}, tan y = \frac{2 t}{1 - t^{2}}$

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Solution

$\begin{matrix} sin x = \frac{2 t}{1 + t^{2}}, tan y = \frac{2 t}{1 - t^{2}} cos x \frac{d x}{d t} = \frac{(1 + t^{2}) \frac{d}{d t} 2 t - 2 t \frac{d}{d t} (1 + t^{2})}{{(1 + t^{2})}^{2}}, s e c^{2} y \frac{d y}{d t} = \frac{(1 - t^{2}) \frac{d}{d t} 2 t - 2 t \frac{d}{d t} (1 - t^{2})}{{(1 - t^{2})}^{2}} cos x \frac{d x}{d t} = \frac{(1 + t^{2}) 2 - 2 t (4 t)}{{(1 + t^{2})}^{2}}, s e c^{2} y \frac{d y}{d t} = \frac{(1 - t^{2}) 2 - 2 t (- 2 t)}{{(1 - t^{2})}^{2}} \frac{d x}{d t} = \frac{(1 + t^{2}) 2 - 8 t^{2}}{{(1 + t^{2})}^{2} cos x}, \frac{d y}{d t} = \frac{(1 - t^{2}) 2 + 4 t^{2}}{{(1 - t^{2})}^{2} s e c^{2} y} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{(1 - t^{2}) 2 + 4 t^{2}}{{(1 - t^{2})}^{2} s e c^{2} y}}{\frac{(1 + t^{2}) 2 - 8 t^{2}}{{(1 + t^{2})}^{2} cos x}} = \frac{((1 - t^{2}) 2 + 4 t^{2}) ({(1 + t^{2})}^{2} cos x)}{({(1 - t^{2})}^{2} s e c^{2} y) ((1 + t^{2}) 2 - 8 t^{2})} \end{matrix}$

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Similar questions

tan y = \frac{2 t}{1 - t^{2}}

sin x = \frac{2 t}{1 + t^{2}}

\frac{d y}{d x}

$s i n x = \frac{2 t}{1 + t^{2}}, t a n y = \frac{2 t}{1 - t^{2}}$

Find $\frac{d y}{d x}$

s i n x = \frac{2 t}{1 + t^{2}}, t a n y = \frac{2 t}{1 - t^{2}},

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| t | < 1

sin x = \frac{2 t}{1 + t^{2}}

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sin x = \frac{2 t}{1 + t^{2}}, tan y = \frac{2 t}{1 - t^{2}}

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